Chapter 8, Problem 78SE

When the population distribution is normal and n is large, the sample standard deviation S has approximately a normal distribution with E(S) ≈ σ and V(S) ≈ σ²/(2n). We already know that in this case, for any n. $\bar{X}$ is normal with E( $\bar{X}$) = μ and V( $\bar{X}$) = σ²/n.

1. a. Assuming that the underlying distribution is normal, what is an approximately unbiased estimator of the 99th percentile θ = μ + 2.33σ?
2. b. When the X_i’s are normal, it can be shown that $\bar{X}$ and S are independent rv’s (one measures location whereas the other measures spread). Use this to compute $V(\hat{\theta })$ and ${\sigma }_{\hat{\theta }}$ for the estimator $\hat{\theta }$ of part (a). What is the estimated standard error ${\hat{\sigma }}_{\hat{\theta }}$?
3. c. Write a test statistic for testing H₀: θ = θ₀ that has approximately a standard normal distribution when H₀ is true. If soil pH is normally distributed in a certain region and 64 soil samples yield $\bar{x}$ = 6.33, s = .16, does this provide strong evidence for concluding that at most 99% of all possible samples would have a pH of less than 6.75? Test using α = .01.

To determine

Find the approximate unbiased estimator of the 99^th percentile of $\theta =\mu +2.33\sigma$.

Answer to Problem 78SE

$\widehat{\theta }=\overline{X}+2.33S$ is the approximate unbiased estimator of $\mu +2.33\sigma$.

Explanation of Solution

Given info:

For a large sample data, the distribution of sample standard deviation is approximately normal. $E\left(S\right)\approx \sigma \text{ and }V\left(S\right)\approx \frac{{\sigma }^2}{2n}$. For any sample size, $\overline{X}$ follows normal distribution if $E\left(\overline{X}\right)=\mu \text{and}V\left(\overline{X}\right)=\frac{{\sigma }^2}{n}$

The underlying distribution is normal. The 99^th percentile is $\theta =\mu +2.33\sigma$.

Calculation:

Unbiased estimator:

If $X_1,X_2\mathrm{.....}{X}_n$ be n random variables with mean $\mu$ and $E\left(\overline{X}\right)=\mu$. Hence, $\left(\overline{X}\right)$ is called unbiased estimator of $\mu$.

Here, the 99^th percentile is $\theta =\mu +2.33\sigma$.

It is known that $E\left(\overline{X}\right)=\mu \text{and}E\left(S\right)\approx \sigma$

Hence,

$\begin{array}{c}E\left(\overline{X}+2.33S\right)=E\left(\overline{X}\right)+2.33E\left(S\right)\\ =\mu +2.33\sigma \end{array}$

Thus, $\widehat{\theta }=\overline{X}+2.33S$ is the approximate unbiased estimator of $\mu +2.33\sigma$.

To determine

Find the value of $V\left(\widehat{\theta }\right),{\sigma }_{\widehat{\theta }}$.

Answer to Problem 78SE

The value of $V\left(\widehat{\theta }\right),{\sigma }_{\widehat{\theta }}$ are $3.7144\times \frac{{\sigma }^2}{n}$ and $\frac{1.9272\times \sigma }{\sqrt{n}}$, respectively.

The standard error is $\frac{1.9272}{\sqrt{n}}\times s$.

Explanation of Solution

Given info:

If $X_i$’s are normal distribution then, $\overline{X}$ and S are independent to each other.

Calculation:

The value of $V\left(\widehat{\theta }\right),{\sigma }_{\widehat{\theta }}$:

It is known that $V\left(S\right)\approx \frac{{\sigma }^2}{2n},V\left(\overline{X}\right)=\frac{{\sigma }^2}{n}$

$\begin{array}{c}V\left(\widehat{\theta }\right)=V\left(\overline{X}+2.33S\right)\\ =V\left(\overline{X}\right)+{2.33}^2\times V\left(S\right),\text{as }\overline{\text{X}}\text{ and S are independent}.\\ =\frac{{\sigma }^2}{n}+5.428\times \frac{{\sigma }^2}{2n}\\ =\frac{{\sigma }^2}{n}\left(1+2.7144\right)\end{array}$

$=3.7144\times \frac{{\sigma }^2}{n}$

$\begin{array}{c}{\sigma }_{\widehat{\theta }}=\sqrt{3.7144\times \frac{{\sigma }^2}{n}}\\ =\frac{1.9272\times \sigma }{\sqrt{n}}\end{array}$

The estimated standard error is

$\begin{array}{c}E\left({\sigma }_{\widehat{\theta }}\right)=E\left(\frac{1.9272\times \sigma }{\sqrt{n}}\right)\\ =\frac{1.9272}{\sqrt{n}}E\left(\sigma \right)\\ =\frac{1.9272}{\sqrt{n}}\times s\end{array}$

Hence, the value of $V\left(\widehat{\theta }\right),{\sigma }_{\widehat{\theta }}$ are $3.7144\times \frac{{\sigma }^2}{n}$ and $\frac{1.9272\times \sigma }{\sqrt{n}}$ respectively.

The standard error is $\frac{1.9272}{\sqrt{n}}\times s$

To determine

Test whether at most 99% of all possible samples would have pH of less than 6.75.

Answer to Problem 78SE

The data does not provide strong evidence for concluding that at most 99% of all possible samples would have a pH of less than 6.75 at 1% level.

Explanation of Solution

Given info:

Soil pH is normally distributed. The sample size is 64. The sample mean and standard deviation are 6.33 and 0.16 respectively. The level of significance is 0.01.

Calculation:

Step 1: Null hypothesis:

$H_0:\mu +2.33\sigma =6.75$

That is, the pH for at most 99% of all possible samples would have pH of 6.75.

Step 3: Alternative hypothesis:

$H_a:\mu +2.33\sigma <75$

That is, the pH for at most 99% of all possible samples would have pH of less than 6.75.

Step 4: Test statistic:

The test statistic is

$\begin{array}{c}z=\frac{\left(\overline{x}+2.33s\right)-6.75}{\frac{1.927s}{\sqrt{n}}}\\ =\frac{\left(6.33+2.33\times 0.16\right)-6.75}{\frac{1.927\times 0.16}{\sqrt{64}}}\\ =\frac{-0.0472}{0.03854}\\ =-1.224\end{array}$

Step 5: P-value:

Software procedure:

Step by step procedure to obtain the P-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘normal’ distribution
• Click the Shaded Area tab.
• Choose X Value and left for the region of the curve to shade.
• Enter the test value as –1.224
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is 0.1105.

Step 6: Decision

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$

Conclusion at $\alpha =0.01$:

Here, the P-value is greater than the level of significance.

That is, $P\text{-value}\left(=0.1105\right)>\alpha \left(0.01\right)$.

By rejection rule, fail to reject null hypothesis.

Step 7: Interpretation:

The data does not provide strong evidence for concluding that at most 99% of all possible samples would have a pH of less than 6.75 at 1% level.